Suppose a ball is thrown
upward from a height of 2 m, with an initial velocity of 15 m/s.

a.Find a formula for the position of the ball as a function of time since it
was thrown.

b.When does it reach its peak?How high does it get?

It reaches its peak when h'(t)=0, ie when v(t)=0.

-9.8t+15=0 -> t=15/9.8 ~1.53 seconds

h(15/9.8)~13.48 m

2.When the brakes are applied to a car, it
undergoes a uniform acceleration of-22 ft/sec2 (this is a
realistic braking force on a dry road – in slippery conditions, the
acceleration will not be as great).Suppose you are driving a car at 60 mi/hr (=88ft/sec), and have to come
to a quick stop.

a. How long will it take you
to come to a complete stop?

b.How
far do you drive in that amount of time?

So you need 176 feet to come to a stop.

4.Use Newton’s
Method to approximate the value of.(What is a good function to use for which this
number will be a root?What would be a
good initial guess?)Apply 2 steps of
this method to get the next 2 improved estimates.

5.A computer monitor shows a picture of a
“growing” rectangle.At a given time, it
is 3 cm long and 2 cm high; the length is growing at a rate of 1 cm/sec, and
the height is growing at a rate of 0.7 cm/sec.At what rate is the area growing?

A = l*h

dA/dt = l*dh/dt + h*dl/dt

=
3cm(0.7cm/s) + 2cm(1cm/sec) = 4.1 cm2/sec

6. Find the point on the
curve nearest to (5,0).

The distance from (x,y) to
(5,0) is .

To simplify computations, I'll minimize d2
instead of d (since one is minimized when the other is, and this avoids the
square root).

Given the constraint, y2=x, so d2
= (x-5)2+x = x2 -9x + 25

This is minimized when 2x-9=0, or x=4.5

So the closest point to this curve is at

7. Maximize and minimize subject to the constraints x+y=5, x,y≥0

I'll replace x with 5-y (I could also solve for y),
and use this to re-write the objective function as a funciton of just y:

So the only cricitial point is when y=10/8=1.25

At this point .Checking the endpoints,

when y=0, f(y) = 25, and y=5, f(y)=75.

So the minimum is 18.75, and this occurs when y=1.2
and x=3.75.The maximum is 75, and this
occurs when x=0 andy=5.

8.Find all critical points of f(x) =
x+25/x.Then find the absolute maximum
and minimum values of this function on the domain [1,10]

f'(x) = 1-25/x2 .

f'(x)=0 if 1=25/x2 , or x2 =25,
so x=-5 or 5

In the given domain, the absolute maximum and minimum
values can only occur at 1,5 or 10, so we check those values:

f(1) = 26, f(5) = 10, f(10) = 12.5

So the maximum is 26, at x=1, and the minimum is 10,
at x=5.

9.F(x) is the antiderivative of the function
shown in this graph.

Find the critical numbers of F, and indicate
whether each is a local maximum, a local minimum, or neither.

The critical numbers of F are when F'(x)=0.But F'(x) is the function graphed, so the
critical numbers of F are at (approximately) 2 and 5

At x=2, F'(x) goes from positive to zero to negative,
so by the first derivative test, this must be a local maximum.

At x=5, F'(x) goes from negative to zero to negative
again, so by the first derivative test, this is not a local extremum.

10. You want to cut a
rectangular beam from a cylindrical log of diameter 12 inches.

The strength of a beam is
proportional to the product of the width and the square of its depth, so to make the beam as strong as possible you
want to maximize this product.

a)Express your
objective as a function of 2 (or more) variables.

If S =
the strength, then S=k(w)(d2), for some constant k

b)What is/are your contraint equation(s)

w2
+ d2 = 122, and w,d>0

c) Solve for the exact
dimensions that will give the beam the greatest strength.

S(w) = k(w)(d2)= k(w)(144-w2)=k(144w-w3)

S'(w)= k(144-3w2), so S'(w) = 0 if
w=Ö48=4Ö3

(I'll
leave it to you to verify that this is in fact the maximum).

At this
point d = Ö(144-48) = Ö96 = 4Ö6, with all measurements in inches.